Solve for x: \dfrac{x^2}{x - 3} = \dfrac{9}{x - 3}. Are any solutions extraneous?

x = -3 is the only valid solution; x = 3 is extraneous Multiply both sides by x - 3: x^2 = 9, so x = \pm 3. But x = 3 makes denominator zero, so it's extraneous. Only x = -3 is valid.

x = -3 is the only valid solution; x = 3 is extraneous

x = 3 is the only valid solution

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Acemy

Nonlinear Equations in One Variable and Systems of Equations

SAT Math· difficulty 5/5

x

A
$No solutions$
B
$x = 3, -3$
C
$x = -3$
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D
$x = 3$

Explanation

x - 3

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